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Suppose two atoms collide (considering one atom to be at rest), in a way such that the energy loss during their collision is more than sufficient to excite the atom (which is at rest) but the remaining energy is not sufficient to excite the first one.

My doubt is, if the energy is completely transfered to the atom at rest, what happens to the remaining energy? Or if the energy is distributed between the atoms, can we find the ratio of energies distributed?

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(a) The title of your question. The concept of coefficient of restitution is useful for macroscopic bodies, but not for microscopic bodies like atoms. The exception is those head-on collisions between atoms when no kinetic energy is lost (elastic collisions). The relative velocity of separation is then equal and opposite to the relative velocity of approach, so we could say that the c of r is 1.

(b) Suppose a helium atom (mass $m_{He}$) moving at velocity $u$ (much less than the speed of light) makes a head-on collision with a stationary sodium atom (mass $m_{Na}$) and excites it, giving it an extra internal energy $\Delta E$.

Using conservation of momentum we have $$m_{He} u = m_{He} v_{He}+m_{Na} v_{Na}$$ in which $v_{He}$ and $v_{Na}$ are the velocities after the collision. But because energy is conserved we also have $$\tfrac{1}{2}m_{He} u^2 = \Delta E + \tfrac{1}{2}m_{He} v_{He}^2+\tfrac{1}{2}m_{Na} v_{Na}^2$$ Provided that $\tfrac{1}{2}m_{He} u^2 > \Delta E$ you can solve these equations for $v_{He}$ and $v_{Na}$.

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Yes, we can find the remaining energy distribution by applying conservation of momentum to the problem. Remembering the energy used to change the energy level of the atom.

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    $\begingroup$ can you please explain by taking a example $\endgroup$
    – IITM
    Jul 18 '20 at 14:15

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